Technically, in (REF ) we further split \(K_{ab}\)  onto a two-dimensional surface (orthogonal to \(u_a\)  of (REF )) extrinsic curvature \(k_{AB}\)  (then \(k_{ab}=\gamma _{aA}\gamma _{bB}k^{AB}\) ) and the surface gravity \(g_{{\cal H}}\) . At this point one may wonder how the extrinsic curvature of the orthogonal to \(u^a\)  2D hypersurface, lying completely inside the 3D hypersurface, is related to the extrinsic curvature tensor \(K_{ab}\) , defining geometry of the hypersurface normal to a vector (\(n^a\)  in the case) directed 
outside. This match happens due to coincidence of \(u^a\)  and \(n^a\)  vectors in the null limit when \(\alpha u^a \rightarrow l^a\)  and \(\alpha n^a \rightarrow l^a\)  at \(\alpha \rightarrow 0\) , where \(\alpha \)  is the renormalization factor needed to keep finiteness of \(K_{ab}\)  components at the event horizon and \(l^a\)  is the null vector normal and tangential to the horizon. The renorm factor \(\alpha \)  is chosen as \(\alpha =f^{1/2}\)  in the case.The requirement for \(\alpha \)  comes as follows [1]}, [2]}. In the \(\alpha \rightarrow 0\)  limit \(\alpha u^a \rightarrow l^a\)  and \(\alpha n^a \rightarrow l^a\) , where \(l^a\)  is a null vector (\(l^a l_a=0\) ) which is tangential and normal to the horizon simultaneously. Clearly, the specific representation of such a null vector depends on the choice. The norm of \(\alpha u^a\)  and \(\alpha n^a\)  of (REF ) with \(\alpha =f^{1/2}\)  is equal to \(f\)  that becomes zero on the horizon.
